令  00=1  ,  fx=∑i=0nix×yi令 \;0^0=1 \; ,\; f_x=\sum\limits^n_{i=0}i^x\times y^i 。

则有：则有 \text{：}

fx+(n+1)x×yn+1  =  ∑i=0n(i+1)n×yi+1+0=∑i=0n[  yi+1(∑j=0xCxj×ij)]=∑i=0n∑j=0xyi+1ijCxj=y×  ∑j=0x(  ∑i=0nij×yi  )=y×  ∑j=0xCxjfj\begin{aligned} f_x+(n+1)^x\times y^{n+1}\;&=\;\sum_{i=0}\limits^n(i+1)^n\times y^{i+1}+0\\ &=\sum_{i=0}\limits^n \bigg[\;y^{i+1}\big( \sum_{j=0}\limits^xC^j_x\times i^j \big) \bigg]\\ &=\sum_{i=0}\limits^n\sum_{j=0}\limits^xy^{i+1}i^jC^j_x\\ &=y\times\; \sum_{j=0}\limits^x \bigg(\; \sum_{i=0}\limits^ni^j\times y^i \; \bigg)\\ &=y\times\; \sum_{j=0}\limits^x C^j_xf_j \end{aligned}

∴fj=  (    (n+1)x×yn+1−y×∑j=0xCxjfj    )  ×  (y−1)−1      [y≠1]\therefore f_j=\; \bigg(\;\; (n+1)^x\times y^{n+1}-y\times \sum_{j=0}\limits^x C^j_xf_j \;\;\bigg)\;\times \;(y-1)^{-1}\;\;\;[y\not =1]

此时，若  y==1  ,则有：此时，若\;y==1\;,则有 \text ：

(n+x)x×yn+1  =  y×∑j=0x−2Cxjfj + y×Cxx−1×fx−1令  x=x+1  ，得：(n+x)x+1×yn+1  =  y×∑j=0x−1Cxjfj + y×Cx+1x×fx进而得：fx  =  [    (n+x)x+1×yn+1 − ∑j=0x−1y Cxjfj    ] ×[ y×Cx+1x ]−1\begin{aligned} (n+x)^x\times y^{n+1}\;&=\;y\times \sum_{j=0}\limits^{x-2} C^j_xf_j\,+\,y\times C^{x-1}_x\times f_{x-1}\\ &令\;x=x+1\;，得\text：\\ (n+x)^{x+1}\times y^{n+1}\;&=\;y\times \sum_{j=0}\limits^{x-1} C^j_xf_j\,+\,y\times C^{x}_{x+1}\times f_x\\ &进而得\text ：\\ f_x\;&=\;\bigg[\;\; (n+x)^{x+1}\times y^{n+1}\,-\,\sum_{j=0}\limits^{x-1} y\,C^j_xf_j \;\;\bigg]\,\times [\, y\times C^x_{x+1} \,]^{-1} \end{aligned}